What's the difference between a compiler's `-O0` option and `-Og` option?

Question

When I want to do debugging of C or C++ programs, I\'ve been taught to use -O0 to turn optimization OFF, and -ggdb to insert symbols into the execu

Answer 1

@kaylum just provided some great insight in their comment under my question! And the key part I really care about the most is this:

[-Og] is a better choice than -O0 for producing debuggable code because some compiler passes that collect debug information are disabled at -O0.

https://gcc.gnu.org/onlinedocs/gcc/Optimize-Options.html#Optimize-Options

So, from now on I'm using -Og (NOT -O0) in addition to -ggdb.

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UDPATE 13 Aug. 2020:

Heck with this! Nevermind. I'm sticking with -O0.

With -Og I get <optimized out> and Can't take address of "var" which isn't an lvalue. errors all over the place! I can't print my variables or examine their memory anymore! Ex:

(gdb) print &angle
Can't take address of "angle" which isn't an lvalue.
(gdb) print angle_fixed_p
$6 = <optimized out>

With -O0, however, everything works fine!

(gdb) print angle
$7 = -1.34869879e+20
(gdb) print &angle
$8 = (float *) 0x7ffffffefbbc
(gdb) x angle
0x8000000000000000:     Cannot access memory at address 0x8000000000000000
(gdb) x &angle
0x7ffffffefbbc: 0xe0e9f642

So, back to using -O0 instead of -Og it is!

Related:

1. [they also recommend -O0, and I concur] What does <value optimized out> mean in gdb?